B. Extinction - ittc.ku.edu

12/3/2007 Exinction notes 1/2

Jim Stiles The Univ. of Kansas Dept. of EECS

B. Extinction Of course, the channel between the transmit and receive antenna is not generally free-space, but instead is filled with all sorts of annoying stuff. Among this stuff are atmospheric gases such as oxygen and water vapor. The gases can surprisingly affect radio wave propagation, particularly at specific frequencies! HO: The Wave Attenuation HO: The Extinction Coefficient HO: Attenuation by Atmospheric Gases In addition to gases, the atmosphere can also be filled with particles—particles such as water droplets (i.e., clouds/fog), snow, dust, and/or rain drops.

12/3/2007 Exinction notes 2/2


Each of these particles can extract energy from a propagating electromagnetic wave. HO: The Extinction Cross-Section If the atmosphere is filled with particles (otherwise know as hydrometeors), we find that a propagating electromagnetic wave likewise exhibits exponential extinction as it propagates! HO: Extinction Due to Hydrometeors

11/29/2007 Wave Attenuation 1/5


Wave Attenuation Recall that a time-harmonic plane-wave solution to the vector wave equation is:

( ) ( ) ( )0j k z tx yˆ ˆr ,t E E e ω−= +E x y

where:

0 0 0kcω ω μ ε= =

This of course describes a wave propagation in free-space, but in reality, e.m. waves on Earth propagate in an atmosphere. Q: Pfft! Aren’t you being a bit picky? The Earth’s atmosphere is practically free-space! A: Nope! From many problems, approximating the atmosphere as free space is accurate. But, for many other problems, it is not at all accurate! The accuracy of the free-space approximation generally depends on three things:

1. The frequency of the propagating wave. 2. The distance over which we are attempting to propagate.



3. What’s occurring in the atmosphere (i.e., the weather conditions!)

The permeability and permittivity of free space is, by definition:

00

1rεε ε εε

= ⇒ = =

Likewise:

00

1rμμ μ μμ

= ⇒ = =

For the Earth’s atmosphere, we find that 1rμ = , but the permittivity of the Earth’s atmosphere is complex:

jε ε ε′ ′′= + where:

{ } { }Re Imε ε ε ε′ ′′= =

Q: Yikes! The dielectric constant of the Earth’s atmosphere is complex?? What the heck does that mean? A: Since the permittivity is complex, so too is our propagation constant k :

( )0 0 r ik j k j kω μ ε ω μ ε ε′ ′′= = + = + where:

{ } { }r ik Re k k Im k= =



And so a plane wave propagating in the Earth’s atmosphere has the form:

( ) ( ) ( )

( ) ( )

( )( ) ( )

r i

ri

ri

j k z tx y

j k j k z j tx y

j k z j tk zx y

j k zy

k z tx

ˆ ˆr ,t E E e

ˆ Ê E e e

ˆ Ê E e

e

e e

ˆ Ê E e

ω

ω

ω

ω

−

+ −

−

−−

−

= +

= +

= +

= +

E x y

x y

x y

x y

Compare and contrast this result with that of free-space:

( ) ( ) ( )0j k z tx yˆ ˆr ,t E E e ω−= +E x y

First, note that the propagating wave in the atmosphere is described using a new term:

ik ze −

Note this is a real-valued function of z ! If we plot the function, we see that it decreases exponentially with respect to z : z

ik ze −



Q: What does this result physically mean? A: Let’s look at the magnitude of this propagating wave:

( ) ( ) ( )22ik z

x yr ,t r ,t r ,t e E E∗ −= ⋅ = +E E E

Note that unlike a plane-wave propagating in free space, the magnitude of a plane-wave in the atmosphere is not a constant throughout all space. Instead, the magnitude of the wave diminishes exponentially as it propagates! Q: But this implies that that something is extracting energy from the propagating wave? A: Exactly! The constituents of the atmosphere (e.g., gases) can absorb electromagnetic energy. This is reflected in a permittivity that is complex—the imaginary part ε ′′ is a result of this fact. In other words, if 0ε ′′ = , then 0ik = and 1 0ik ze .− = . No attenuation occurs! The second thing to note when comparing the two plane-wave expressions is the phase terms: ( )0j k z te ω− for a plane wave in free-space



( )rj k z te ω− for a plane wave in the atmosphere Thus, the value rk is the spatial frequency (radians/meter) of a plane wave propagating in the atmosphere. From this value we can find the propagation wavelength:

2 2r

rk

kπ πλλ

= ⇒ =

and the propagation velocity:

r pp r

k vv kω ω

= ⇒ =

We find that the velocity of a plane-wave in our atmosphere will be less than that of a plane-wave in free space ( pv c≤ ).

12/3/2007 The Extinction Coefficient 1/3


The Extinction Coefficient Q: So what’s the power density of a plane-wave propagating in the Earth’s atmosphere? A: Recall the magnitude of the power density of a propagating plane wave was found from the Poynting vector:

( ) 2

2rη

=E

W

And so the power density of a plane-wave propagating in the atmosphere is:

222

22

2

2

2

i

i

k zx y

x y k z

e E E

E Ee

η

η

−

−

⎛ ⎞+⎜ ⎟⎝ ⎠=

+=

W

And so (not surprisingly), we find that the power density diminishes exponentially as:

2 ik ze −∝W We now define the extinction coefficient eκ :



2 ie kκ So that now:

e ze κ−∝W It is apparent that the extinction coefficient describes the “rate” at which the power density exponential decays.

0 no attenuation of waveeκ = ⇒

0 exponential reduction of waveeκ > ⇒

Note that the extinction coefficient is a bit of a misnomer. The term ezκ is unitless, and so the extinction “coefficient”

eκ must have units of inverse length (e.g., 1m − or 1km − ). However, we typically express the extinction coefficient in terms of “dB/km”, as in (for example):

2 3edB.km

κ =

Q: Yikes! Why would we do that? What does it mean? A: Consider the case where a plane wave propagates a distance , beginning at location z and ending at location z + .



The attenuation (i.e., loss) of the plane wave power density over this distance can be described as:

( )( )

( )e e ee

e e

z z

z zz e e e e

z e e

κ κ κκ

κ κ

− + − −−

− −

+= = =

WW

Expressed in decibels:

( )( )

( )( )

10

10

110

Attenuation in dB 10

10

104 343

e

e

e

zlogz

log e

log e.

κ

κκ

−

⎛ ⎞+= − ⎜ ⎟⎜ ⎟

⎝ ⎠= −

=

=

WW

And so, we can define an “attenuation in dB/unit length” as:

Attenuation in dB 4 343edB.km

κ=

12/3/2007 Attenuation by Atmospheric Gasses 1/4


Attenuation by Atmospheric Gasses

Q: So, what exactly is the value of the extinction coefficient

eκ of the Earth’s atmosphere? A: This value depends on many things, including altitude, temperature, pressure, and weather conditions. But mostly, it depends on the frequency of the propagating signal! For a “clear” atmosphere (i.e., clear weather), we find that the extinction coefficient increases dramatically with frequency.



For example, a propagating wave in the Earth’s atmosphere with a frequency of 10 GHz is attenuated an additional 0.03 dB for every km it propagates (i.e., 0 3e . dB kmκ = ). Thus, if this wave propagates over a distance of 50 km, it is attenuated by an additional 1.5 dB. If, however, the signal of this propagating wave is increased to 60 GHz, we find that atmospheric attenuation has increased dramatically, to approximately 20.0 dB for every km it propagates (i.e., 20 0e . dB kmκ = ). Thus, if this wave propagates over a distance 50 km, it is attenuated by an additional 1000 dB !!!!! Thus, we see that below 10 GHz the attenuation due to atmospheric gasses is largely regarded as insignificant. Conversely, above 50 GHz the attenuation is so large that it make radio wave propagation over long distances effectively impossible!

Remember, atmospheric attenuation limits the “useable” region of the electromagnetic spectrum!

Q: Why does the atmospheric attenuation “peak” at several frequencies? A: Quantum mechanics! Molecules and atoms absorb energy by discrete energy values (i.e., quanta). For each energy level

nE ( 1 2 3n , , ,= ), there is a corresponding wave frequency, a frequency that can be determined from Plank’s Constant :



n

n n nEE f f= ⇒ =

It turns out that both a water molecule ( 2H O ) and oxygen molecule ( 2O ) exhibit an number of these absorption frequencies in the microwave region. For example, we find that water vapor absorbs e.m. energy at about 22 GHz, whereas oxygen absorbs e.m. energy at about 60 GHz. Q: Now, in your example you said that the propagating wave is:

“attenuated by an additional 1000 dB!” What did you mean by “additional” attenuation?



A: Remember, the power density of a spherical wave produced by an antenna diminishes as 21 r :

( ) ( ) 2

ˆr U ,r

θ φ=rW

Thus, the attenuation by the atmosphere is in addition to this

2r − “spreading loss”. A more accurate description of the power density of a spherical wave is:

( ) ( ) 2

ere ˆr U ,r

κθ φ

−

=W r

And so, a more accurate form of the Friis Transmission Equation is likewise:

024

eremr A

G AP P er

κπ

−=

12/3/2007 The Extinction Cross Section 1/7


( )i rW Pa

Ps

( ∆ ) ( )i ir r r+ <W W

The Extinction Cross-Section

Say a lossy particle is illuminated by a plane wave with power density ( )i rW : The particle will: 1. Absorb energy at a rate Pa Watts. 2. Scatter energy (in all directions) at a rate Ps Watts. Because of conservation of energy, the power density of the incident wave must be diminished after interacting with the particle:

( )i rW



We can therefore define the absorption cross-section of this particle as:

( )2a

ai

Pσ mr

⎡ ⎤⎣ ⎦W

and likewise the total scattering cross-section as:

( )2s

si

Pσ mr

⎡ ⎤⎣ ⎦W

Note that using these definitions we find that:

( ) ( )anda a i s s iP σ r P σ r= =W W

We can likewise define a particle extinction cross-section as:

( ) ( ) ( )2a s a s

e a si i i

P P P Pσ σ σ mr r r

+ ⎡ ⎤= + = + ⎣ ⎦W W W



The extinction cross-section (as well as aσ and sσ ), depend on the physical properties (size, shape, material, orientation) of the particle, as well as the frequency of the incident wave. Q: So what is the extinction cross-section of say, a rain drop or a dust particle, or a snow flake? A: A difficult question to answer, for two reasons:

1. Every rain drop/dust particle/snow flake is different, so we can only determine the extinction cross-section of a typical (i.e., average) particle. 2. Determining the extinction cross-section of even a typical particle is extremely difficult!

Evaluating the extinction cross-section of a particle requires an extremely complicated and difficult electromagnetic analysis (the solution must satisfy Maxwell’s equations!). In fact, solutions exist for only the simplest of objects; for example, a perfect sphere! Even then, the solution for the extinction cross-section of a simple sphere is quite complex, and is expressed as a infinite series solution known as the Mie Series:

( ) ( ) ( )2 2 40 1 0 2 0

1

ne en e e

nk a k a k aσ σ σ σ

∞

=

= = + +∑



where the values enσ are the coefficients of the Mie Series, and the value a is the radius of the sphere. Thus, the quantity 0k a is simply the electrical radius of the sphere—the radius of the sphere expressed in radians:

02 2 ak a aπ πλ λ

⎛ ⎞= =⎜ ⎟⎝ ⎠

Note 0k a is very much like the value β we studied in transmission line theory (i.e., the electrical length of a transmission line, expressed in radians). Q: Yikes! What good is an infinite series solution? Wouldn’t this require an infinite number of calculations? A: To get an exact answer yes. But we can truncate the infinite series (making it finite) and still get a very accurate (albeit imperfect) approximation. For example, if we retained the first three terms:

( ) ( ) ( )2 4 61 0 2 0 3 0e e e ek a k a k aσ σ σ σ≈ + +

Q: But how many terms should we retain? How do we know when to truncate the series? A: Note that as the sphere gets electrically small (i.e., 0 1k a ) the higher order terms (e.g., ( )4

0k a or ( )120k a ) get really small, and thus can be ignored.



Conversely then, as the electrical size of the sphere increases, more and more terms of the Mie Series will become significant. Q: So how does all this help us determine the extinction cross-section of a non-spherical particle such as raindrop or dust particle? A: Note that your average raindrop has a diameter of a few millimeters, whereas the wavelength of a propagating wave at, say, 3.0 GHz is 100 mm. Thus, your average hydrometer is electrically small (i.e., 0 1k a ) in the RF/Microwave region of the electromagnetic spectrum. From the standpoint of the Mie Series, this means that only a few terms will be significant. We find for this special case that the Mie Series provides us with a relatively simple answer:

( )40

4

0

183 2

18 13 2

re p

r

rp

r

k a A

a A for k ac

εσε

εωε

⎛ ⎞−≈ ⎜ ⎟+⎝ ⎠

⎛ ⎞−⎛ ⎞= ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

where rε describes the material of the sphere, and pA is its physical cross-section:

2pA aπ=



This result is a quite famous one, known as the Rayleigh solution. Q: Terrific. But again, how does the solution for a perfect sphere help us determine the extinction cross-section of non-spherical particles such as raindrops or snowflakes!?! A: It turns out the Rayleigh solution for a sphere is a good approximation for electrically small non-spherical particles as well! We can define an equivalent physical cross-section pA and thus an equivalent radius a for any non-spherical particle. Inserting these equivalent values into the Rayleigh solution for spherical particles provides an “acceptably” accurate value of the particle extinction cross-section—provided of course that the non-spherical particle is electrically small! In fact, all electrically small particles are known as Rayleigh Scatterers.

a Particle Equivalent

Physical Cross Section



Of course, what makes a particle electrically small is not just its physical size, but instead its physical size with respect to a wavelength. Thus, the frequency of the propagating wave makes a BIG difference! We can see this in the Rayleigh solution of extinction cross-section:

4

018 1

3 2r

e pr

a A for k ac

εωσε

⎛ ⎞−⎛ ⎞≈ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

Note that the extinction cross-section is proportional to 4ω . Thus, if we double the frequency of the propagating wave, the extinction cross-section will increase 8 times (providing the particle is still electrically small)! We will find that this make propagating through rain or fog exceeding difficult at high frequencies!

λ

Dad-gum Rayleigh Scatterers!

12/3/2007 Extinction due to Hydrometeors 1/9


Extinction Due to Hydrometeors

Consider a plane wave that propagates a distance ∆x through a thin volume with cross-sectional area A.

( )xW

x̂

Volume of region is

∆V A x=

x̂

( ∆ )x x+W

∆x

Area A



Say that this volume is filled with N particles. The particle density no is therefore defined as:

3∆oparticlesN Nn

V A x m⎡ ⎤

= ⎢ ⎥⎣ ⎦

Now, we know that energy is flowing into the front of this volume at a rate of:

( ) [ ]inP x A W= W

while the power of the plane wave exiting the back of the volume is:

( ) [ ]∆inP x x A W= +W

Of course, the particles within the volume will extract power from this plane wave (due to absorption and scattering). Thus, the power lost due to extinction can be written as:

( ) ( )∆

∆

∆

out inP P Px x A x A

A

= −

= + −

=

W W

W

Where ( ) ( )∆ ∆x x x+ −W W W . Note since out inP P< , both ∆P and ∆W will be negative values (i.e., the power decreases as it passes through the volume).



Q: What happened to the missing energy ? A: The particles within the volume extract this energy by absorbing or scattering the incident plane wave. If enP is the rate at which energy is extracted by the n-th particle, then the total rate of energy extinction by the entire collection is:

( )

1

1

N

e ennN

enn

P P

σ x

=

=

=

=

∑

∑ W

By conservation of energy, we can conclude that:

∆e in outP P P P= − = −

And therefore:

( )1

∆ eN

enn

P P

σ x=

= −

= −∑ W

Q: Yikes! How are we supposed to know eσ for each and every one of the N particles? A: Look closer at the equation! We don’t need to all the values enσ --we simply need to know the sum of all the values

(i.e., 1

N

enn

σ=∑ )!



To determine this sum, we just need to know the average value of the extinction cross-sections ( eσ ), defined as:

1

1 N

e enn

σ σN =∑

Therefore:

( )

( )

( ) ( )( )

1

1

∆N

enn

N

enn

e

e

P σ x

x σ

x N σN x σ

=

=

= −

= −

= −

= −

∑

∑

W

W

WW

Recall, however, that:

∆o oN n V n A x= =

Therefore: ( )( )

∆∆

e

o e

P N x σn A x σ x

= −

= −

WW

and thus:

( )∆∆ ∆o eP n x σ x

A= = −W W

Finally (whew!) we can say:

( )∆∆ o en x σ

x= −

WW



Extinction due to particles

Extinction due to gases.

And taking the limit as ∆ 0x → , we have determined the following differential equation:

( ) ( ) ( )∆ 0

∆∆ o ex

d xlim n σ xx dx→

= = −WW

W

This differential equation is easily solved:

( ) ( ) ( )0 o en σ xx x e −= =W W

And thus: ( ) ( ) ( )0 o en σ x ˆx x e −= =W W x

Note the similarity of the function

( )o en σ xe − with exe κ− It is apparent that the extinction due to atmospheric particles has precisely the same form as the extinction due to atmospheric gases! In fact, we find that the extinction coefficient for atmospheric particles (such as hydrometeors) is:



e o en σκ

As a result, we can describe the power density of a plane wave propagating in the atmosphere as:

( ) ( ) 2

ere ˆr U ,r

κθ φ

−

=W r

Where the extinction coefficient eκ is due to gasses, particles, or both! Now, consider the (likely) situation where the atmospheric particles are Rayleigh scatterers. Since the extinction coefficient is proportional the average particle extinction cross-section:

e eσκ ∝

And since the extinction cross-section of a Rayleigh scatterer is proportional to frequency as:

4eσ f∝

We can likewise conclude that the extinction coefficient is proportional to frequency as:

4e fκ ∝



Thus, the extinction coefficient of a volume of Rayleigh Scatterers is proportional to the forth power of the wave frequency. If we double the frequency, we increase the extinction coefficient 8 times! This is unfathomably significant, as we know that the power density of a propagating wave is exponentially related to the extinction coefficient (i.e., ere κ−∝W ). We find that as signal frequency increases, it rapidly becomes nearly impossible to propagate through an average rainstorm!

Extinction Coefficient as a function of frequency for various rainfall rates

Extinction Coefficient as a function of frequency for various densities of fog/clouds



For example, using the charts above, the extinction coefficient for an atmosphere containing rain falling at a rate of 1 0. inches/hour (i.e., 25 mm/hour), is approximately:

0.01 dB/km at 3.0 GHz





Thus, if this rainstorm extends over 10 km, there will be an additional wave attenuation of: 0.1 dB at 3.0 GHz 0.8 dB at 6.0 GHz 6.0 dB at 12.0 GHz 35 dB at 24.0 GHz 80 dB at 48.0 GHz

Note the big difference between 3.0 GHz and 48.0 GHz!

10 km



Q: I notice the effect of Rayleigh Scattering when the frequency is low. When we double the frequency from 3.0 to 6.0 GHz, the extinction increases 8 times from 0.01 dB/km to 0.08 dB/km. Again, when we go from 6.0 GHz to 12.0 GHz, the extinction increases approximately 8 times to 6 dB/km. But, when we increase the frequency to 24.0 GHz and then to 48.0 GHz, this “ 4f “ Rayliegh Scattering effect seems to disappear—the extinction does not increase 8 times?! A: Remember, as the wave frequency increases, the electrical size of the raindrops also increase. If the frequency becomes too high, then the particles will no longer be electrically small (i.e., 0 1k is not true). As a result, other terms in the Mie Series become significant, and the extinction no longer increases as 4f . In fact, the extinction coefficient eventually “plateaus” as frequency is increased—the Mie Series has converged!

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